$\dfrac{d}{dx}\left(\dfrac5x \ln(x)\right)=$
Answer: $\dfrac5x \ln(x)$ is the product of two, more basic, expressions: $\dfrac5x$ and $\ln(x)$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac5x \ln(x)\right) \\\\ &=\dfrac{d}{dx}\left(\dfrac5x\right)\ln(x)+\dfrac5x\dfrac{d}{dx}(\ln(x))&&\gray{\text{The product rule}} \\\\ &=5\dfrac{d}{dx}\left(\dfrac1x\right)\ln(x)+\dfrac5x\dfrac{d}{dx}(\ln(x))&&\gray{\text{Constant multiple rule}} \\\\ &=5\dfrac{d}{dx}\left(x^{-1}\right)\ln(x)+\dfrac5x\dfrac{d}{dx}(\ln(x))&&\gray{\text{Rewrite }\dfrac1x\text{as a power}} \\\\ &=5\cdot (-1x^{-2})\cdot\ln(x)+\dfrac5x\cdot \dfrac1x&&\gray{\text{Differentiate }x^{-1}\text{ and }\ln(x)} \\\\ &=-5x^{-2}\ln(x)+\dfrac{5}{x^2}&&\gray{\text{Simplify}} \\\\ &=\dfrac{5-5\ln(x)}{x^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac5x \ln(x)\right)=\dfrac{5-5\ln(x)}{x^2}$ or any other equivalent form.